Is this a sufficient statistic for uniform distribution?

Question

I'm having trouble proving that given a uniform distribution $X_i \sim U(0,\theta)$ with $\theta$ unknown, the statistics $2\bar X$ and $\bar X$ are not sufficient; any ideas?

Thanks for your help.

Answer 1

How to spot an insufficient statistic.

Suppose that $u({\bf y})$ is a function of the data vector ${\bf y}=(y_1,\dots, y_n).$ The factorization theorem says that if $u$ is sufficient for $\theta$, then for any two parameters $\theta$ and $\theta^*$, and any two data vectors ${\bf y}$ and ${\bf y^*}$ with $u({\bf y})=u({\bf y^*})$, we have $L(\theta,{\bf y})\,L(\theta^*,{\bf y^*})= L(\theta,{\bf y^*})\,L(\theta^*,{\bf y}).$

In your problem, provided $n>1$, it is easy to find parameters and data vectors so that this equation fails.

Answer 2

Here's a hint: $ \operatorname{E}(2\bar X) = \theta,$ so $2\bar X$ is an unbiased estimator of $\theta$, but if, for example, $(X_1,X_2,X_3) = (1,2,12)$ then the estimate of $\theta$ is actually smaller than the largest of the three observations. That shows there is more information about $\theta$ in the sample than there is in $\bar X$.

Here's somewhat more than a hint: The conditional distribution of $X_1,X_2,X_3$ given that $\bar X = 1$ and $\theta = 2$ is supported in some subset of $[0,2]^3$, but the conditional distribution of $X_1,X_2,X_3$ given that $\bar X = 1$ and $\theta=20$ has the point $(0,0,3)$ in its support. (Note that $(0+0+3)/3 = \bar X = 1$.) That shows that $\bar X$ is not sufficient without helping you find any one-dimensional sufficient statistic. In fact, one exists.